什么是重新present任何双重价值需要的字符的最大长度？ [英] What is the maximum length in chars needed to represent any double value?

问题描述

当我转换成一个无符号的8位int字符串，然后我知道结果永远是最多3个字符（255）和一个符号的8位int，我们需要4个字符为例如-128。

When I convert an unsigned 8-bit int to string then I know the result will always be at most 3 chars (for 255) and for an signed 8-bit int we need 4 chars for e.g. "-128".

现在什么，我其实不知道是浮点值同样的事情。什么是重新present所需字符的最大数量的双或浮动值作为一个字符串？

Now what I'm actually wondering is the same thing for floating-point values. What is the maximum number of chars required to represent any "double" or "float" value as a string?

假设一个普通的C / C ++双（IEEE 754）和正常的十进制扩展（即未％E的printf格式）。

Assume a regular C/C++ double (IEEE 754) and normal decimal expansion (i.e. no %e printf-formatting).

我甚至不知道，如果真的小数目（即0.234234）会比真正庞大的数字（双打重新presenting整数）？再

I'm not even sure if the really small number (i.e. 0.234234) will be longer than the really huge numbers (doubles representing integers)?

推荐答案

标准头＆LT; FLOAT.H＆GT; 在C或＆LT ; cfloat＆GT; 在C ++中，包含了几个常数做的范围和浮点类型的其他指标。其中之一是 DBL_MAX_10_EXP ，最大功率的-10指数重新present所有双击值需要。由于 1EN 需要 N + 1 数字重新present，并有可能是一个负号为好，那么答案

The standard header <float.h> in C, or <cfloat> in C++, contains several constants to do with the range and other metrics of the floating point types. One of these is DBL_MAX_10_EXP, the largest power-of-10 exponent needed to represent all double values. Since 1eN needs N+1 digits to represent, and there might be a negative sign as well, then the answer is

int max_digits = DBL_MAX_10_EXP + 2;

此假定指数比的重新present最大可能尾数值所需的位数大;否则，也将有一个小数点后跟多个数字。

This assumes that the exponent is larger than the number of digits needed to represent the largest possible mantissa value; otherwise, there will also be a decimal point followed by more digits.

修正

最长的数量实际上是最小的重presentable负数：需要足够的数字来涵盖指数和尾数。此值 -pow（2 DBL_MIN_EXP - DBL_MANT_DIG），其中 DBL_MIN_EXP 为负。这是很容易看到（和通过感应证明）的 -pow（2 -N）需求 3 + N 对于非科学小数重presentation字符（ - 0，后跟 N 位数） 。因此，答案是

The longest number is actually the smallest representable negative number: it needs enough digits to cover both the exponent and the mantissa. This value is -pow(2, DBL_MIN_EXP - DBL_MANT_DIG), where DBL_MIN_EXP is negative. It's fairly easy to see (and prove by induction) that -pow(2,-N) needs 3+N characters for a non-scientific decimal representation ("-0.", followed by N digits). So the answer is

int max_digits = 3 + DBL_MANT_DIG - DBL_MIN_EXP

对于64位IEEE双，我们有

For a 64-bit IEEE double, we have

DBL_MANT_DIG = 53
DBL_MIN_EXP = -1023
max_digits = 3 + 53 - (-1023) = 1079

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